Determine the values of k for which the function f(x)=4x^2-3x+2kx+1 has 2 zeros
recall that the formula for discriminant:
D = b^2 - 4ac
if
D < 0 : two imaginary roots
D = 0 : one root
D > 0 : two real roots
thus, given the equation, we can substitute the values of a, b and c in the discriminant, which must be > 0:
(2k - 3)^2 - 4(4)(1) > 0
4k^2 - 12k + 9 - 16 > 0
4k^2 - 12k - 7 > 0
(2k - 7)(2k + 1) > 0
these are the ranges of values:
k > 7/2
k < -1/2
hope this helps~ :)
To find the values of k for which the function f(x) has 2 zeros, we need to set the discriminant of the quadratic equation equal to zero. The discriminant of the quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac.
In this case, the quadratic equation is 4x^2 - 3x + 2kx + 1 = 0. Let's rearrange the terms to resemble the standard form ax^2 + bx + c = 0:
4x^2 - (3 - 2k)x + 1 = 0.
Now we can calculate the discriminant:
(3 - 2k)^2 - 4(4)(1) = 0.
Simplifying the equation:
9 - 12k + 4k^2 - 16 = 0.
Rearranging terms and combining like terms:
4k^2 - 12k - 7 = 0.
Now we have a quadratic equation in terms of k. We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
k = (-b ± √(b^2 - 4ac)) / (2a),
where a = 4, b = -12, and c = -7:
k = (-(-12) ± √((-12)^2 - 4(4)(-7))) / (2(4)).
Simplifying the equation:
k = (12 ± √(144 + 112)) / 8,
k = (12 ± √256) / 8,
k = (12 ± 16) / 8.
This gives us two possible values for k:
k1 = (12 + 16) / 8 = 28 / 8 = 3.5,
k2 = (12 - 16) / 8 = -4 / 8 = -0.5.
Therefore, the function f(x) = 4x^2 - 3x + 2kx + 1 has two zeros for k = 3.5 and k = -0.5.